I have solved this by finding Discriminant.

and getting $x = -2a, \;\text {and} \;2a$

Is this the right method?

and getting $x = -2a, \;\text {and} \;2a$

Is this the right method?

0 votes

If $x$ is real, the set of real values of $a$ for which the function $$y=x^2-ax+1-2a^2$$ is always greater than zero is

- $- \frac{2}{3} < a \leq \frac{2}{3}$
- $- \frac{2}{3} \leq a < \frac{2}{3}$
- $- \frac{2}{3} < a < \frac{2}{3}$
- None of these

1 vote

Refer this on how to use discriminant to solve such problems.

For the quadratic function $ax^{2}+bx+c=0$ to have positive value, the discriminant $b^{2}-4ac<$ $0$

So, on simplifying for the given function, we get $9$$a^{2}$ $<$ $4$ $\Rightarrow$ $a^{2}$ $<$ $\frac{4}{9}$ $\Rightarrow$ $-\frac{2}{3}$ $<$ $a$ $<$ $\frac{2}{3}$

Option C is the answer.